SOLUTIONS OF DIFFERENTIAL EQUATIONS OF THE FIRST ORDER AND FIRST DEGREE :

(A) Differential equation of the form dy_dx = f(x) or dy

dx = f(y)

Integrate both sides i.e.

z

dy =

z

f x( ) dx or

z

dy =

z

dx to get its solution.

(B) Variable Separable Form : Differential equation of the form dy_dx = f(x) g(y)

This can be integrated as

dy g y( )

z

=

z

f(x) dx + c

(C) Homogeneous Equations : It is a differential equation of the form dy_dx = f x y_{g x y}( , )_{( , )}, where f(x, y) and g(x, y) are homogeneous functions of x and y of the same degree. A function f(x, y) is said to be homogeneous of degree n if it can be written as xn _f y

x

F

HG

IKJ

or yn _f x y

F

HG

IKJ

. Such an equation can be solved by putting y = vx or x = vy. After substituting y = vx or x = vy. The given equation will have variables separable in v and x. (D) Equations Reducible to Homogeneous form and

variable separable form

* Form dy_dx = _{Ax By C}ax by c₊+ +₊ ... (1) where _Aa ≠ _Bb

This is non Homogeneous

Put x = X + h and y = Y + k in (1)

∴ dy_dx = dY_dX Put ah + bk + c = 0, Ah + Bk + C = 0, find h, k

Then dY_dX = aX bY_{AX BY}+₊ . This is homogeneous.

* Form dy dx = ax by c Ax By C + + + + ... (1), where _Aa = b_B = k say ∴ dy_dx = k_{Ax By C}(Ax₊+By₊)+c Put Ax + By = z ⇒ A + B dy_dx = dz_dx ⇒ dz_dx = A + B kz c_{z c+}+

This is variable separable form and can be solved. * Form dy_dx = f(ax + by + c)

Put ax + by = z ⇒ a + b dy_dx = _dxdz

∴ dz_dx = a + b f(z)

This is variable separable form and can be solved. (E) Linear equation :

* In y : dy_dx + Py = Q, where P, Q are function of x alone or constant.

its solution ye

z

P dx =

z

Q e

z

P dx dx + c

where _e

z

P dx _{is called the integrating factor (I.F.) of} the equation.

* In x : dx_dy + Rx = S, where R, S are functions of y alone or constant.

its solution xe

z

R dy =

z

S e.

z

R dy. dy + c

where _e

z

R dy. _{is called the integrating factor (I.F.) of} the equation.

(F) Equation reducible to linear form :

* Differential equation of the form dy_dx + Py = Qyn where P and Q are functions of x or constant is called Bernoulli's equation. On dividing through out by yn_{, we} get

y–n dy

dx + py–n + 1 = Q

Put y–n + 1 _{= z}

⇒ The given equation will be linear in z and can be solved in the usual manner.

Note : In general solution of differential equation we can take integrating constant c as tan–1 _{c, e}c_{, log c etc.} according to our convenience.

VECTORS

1. Types of vectors :

(a) Zero or null vector : A vector whose magnitude is zero is called zero or null vector.

(b) Unit vector : $a = r a a | | = Vector a Magnitude of a

(c) Equal vector : Two vectors a and _b are said to be equal if |a| = |b| and they have the same direction. 2. Triangle law of addition : AB + BC = AC

c = a + b c = a + b – – – a b – – A C B

3. Parallelogram law of addition : OA + OB = OC a + b = c a– b– C B A D

where OC is a diagonal of the parallelogram OABC

4. Vectors in terms of position vectors of end points -

AB = OB – OA = Position vector of B – position vector of A

i.e. any vector = p.v. of terminal pt – p.v. of initial pt. 5. Multiplication of a vector by a scalar :

If ar is a vector and m is a scalar, then mar is a vector and magnitude of mar = m|a|

and if ar = a1$i + a2$j + a3$k

then mar = (ma1)$i + (ma2)$j + (ma3)$k 6. Distance between two points :

Distance between points A(x1, y1, z1) and B(x2, y2, z2) = Magnitude of AB→

= (x2−x1)2+(y2−y1)2+(z2−z1)2

7. Position vector of a dividing point :

(i) If A(a) & B(br) be two distinct pts, the p.v. cof the point C dividing [AB] in ratio m1 : m2 is given by

r c = m b m a_m1 _m2 1 2 r _r + +

(ii) p.v. of the mid point of [AB] is 1₂ [p.v. of A + p.v. of B] (iii) If point C divides AB in the ratio m1 : m2 externally,

then p.v. of C is _c = m b m a_m1 _m2

1 2

− −

10. Coplanar and non coplanar vector :

(i) If a, b, c be three non coplanar non zero vector then xa + yb + zc = 0

⇒ x = 0, y = 0, z = 0

(ii) If a, _b, c be three coplanar vectors, then a vector

c can be expressed uniquely as linear combination of remaining two vectors i.e. c = λa + µb

(iii) Any vector _r can be expressed uniquely as inner com- bination of three non coplanar & non zero vectors a,

b and c i.e. _r = xa + yb + zc

11. Products of vectors :

(I) Scalar or dot product of two vectors : (i) a . b = |a| |b| cosθ

(ii) Projection of a in the direction of b = a b_{| |b}. & Projection of b in the direction of a = a b_{| |}._a (iii) Component of _r on a =

FHG

_{| |}r a_a.2

IKJ

a

Component of _r ⊥ to a = _r –

FHG

_{| |}r a_a.2

IKJ

a (iv) $i.$i = $j.$j = $k.$k = 1

(v) $i.$j = $j.$k = $k.$i = 0 (iv) p.v. of centriod of triangle formed by the points A(a),

B(br) and C (cr) is a b c+ +

3

(v) p.v. of the incentre of the triangle formed by the points A(αr), B(βr) and C(rγ) is

a b c

a b c

α+ β+ γ

+ + where a = |BC|, b = |CA|, c = |AB|

8. Some results :

(i) If D, E, F are the mid points of sides BC, CA & AB respectively, then AD + _BE + _CF = ₀

(ii) If G is the centriod of ∆ABC, then GA + GB + GC = 0

(iii) If O is the circumcentre of a ∆ABC, then

OA + OB + OC = 3 OG = OHwhere G is centriod and H is orthocentre of ∆ABC.

(iv) If H is orthocentre of ∆ABC, then

HA + HB + HC = 3HG = OH

9. Collinearity of three points :

(i) Three points A, B and C are collinear if _AB = λ_AC for some non zero scalar λ.

(ii) The necessary and sufficient condition for three points with p.v. a, _b, c to be collinear is that there exist three scalars l, m, n all non zero such that

(vi) If a and _b are like vectors, then a._b = |a||_b| and If a and b are unlike vectors, then a.b = –|a||b| (vii) a,b are ⊥ ⇔ a.b = 0

(viii) (a.b).b is not defined (ix) (a ± b)2 _{= a}2 _{± 2a.}

b + b2

(x) |a + b| = |a| + |b| ⇒ a || b

(xi) |a + b|2 _{= |a|}2 _{+ |b|}2 _{⇒ a ⊥ b} (xii) |a + b| = |a – b| ⇒ a ⊥ b (xiii) work done by the force :

work done = F.d, where F is force vector and d

is displacement vector.

(II) Vector or cross product of two vectors : (i) a × b = |a| |b| sinθ $n

(ii) if a, b are parallel ⇔ a × b = 0 (iii) a × _b = –(_b × a)

(iv) $n = ± _|a b_{a b}×_{× |}

(v) let a = a1$i+ a2$j + a3$k & b = b1$i+ b2$j + b3$k, then

a × b = $ $ $ i j k a a a b b b 1 2 3 1 2 3 (vi) a × a = 0 (vii) a × (_b × c) = (a × _b) × c (viii) a × (b + c) = (a × b ) + (a × c) (ix) $i × $i = $j × $j = $k × $k = 0, $i × $j = $k, $j × $k = $i, $k × $i = $j (x) Area of triangle : (a) 1₂ AB AC×

(b) If a, b, care p.v. of vertices of ∆ABC, then = 1₂ |(a × b) + (b × c) + (c × a)| (xi) Area of parallelogram :

(a) If a & _bare two adjacent sides of a parallelo- gram, then area = |a × b|

(b) If a and b are two diagonals of a parallelogram, then area = 1₂ |a × b|

(xii) Moment of Force :

Moment of the force F acting at a point A about O is Moment of force = _OA ×_F = r × _F

(xiii) Lagrange's identity : |a × b|2 ₌ a a a b

a b b b.. ..

(III) Scalar triple product : (i) If ar = a1$i + a2$j + a3$k,

r

b = b1$i + b2$j + b3$k and r

(ar × br).cr = [ar br cr] = a a a b b b c c c 1 2 3 1 2 3 1 2 3

and [ar br cr] = volume of the parallelopiped whose coterminus edges are formed by ar, br, cr

(ii) [ar br cr] = [br cr ar] = [cr arbr],

but [ar br cr] = – [br ar cr] = – [ar cr br] etc.

(iii) [ar br cr] = 0 if any two of the three vectors ar, br,

r

c are collinear or equal.

(iv) (ar × br).cr = ar.(br × cr) etc. (v) [$i $j $k] = 1

(vi) If λ is a scalar, then [λarbr cr] = λ[ar br cr] (vii) [ar + dr br cr] = [ar br cr] + [drbrcr] (viii) ar, br, cr are coplanar ⇔ [ar br cr] = 0

(ix) Volume of tetrahedron ABCD is 1₆|AB→ × AC→ . AD→ | (x) Four points with p.v. ar, br, cr, dr will be coplanar if

[dr br cr] + [drcr ar] + [drarbr] = [ar br cr] (xi) Four points A, B, C, D are coplanar if

[AB→ AC→ AD→ ] = 0

(xii) (a) [ar + br br + cr cr + ar] = 2[ar br cr] (b) [ar – br br – cr cr – ar] = 0

(c) [ar × br br × cr cr × ar] = [ar br cr]2

(d) If ar, br, cr are coplanar, then so are ar × br,

r

b × cr, cr × ar and r

a + br, br + cr, cr + ar and ar – br, br – cr, r

c – ar are also coplanar. (IV) Vector triple Product :

If ar, br, cr be any three vectors, then (ar × br) × cr

and ar × (br × cr) are known as vector triple product and is defined as

(ar × br) × cr = (ar.cr)br – (br.cr)ar and ar × (br × cr) = (ar.cr)br – (ar.br)cr

Clearly in general ar × (br × cr) ≠ (ar × br) × cr but (ar × br) × cr = ar × (br × cr) if and only if ar, br

& cr are collinear

12. Application of Vector in Geometry :

(i) Direction cosines of r ai bj ckr = $+ $+ $ are _{| |}a_rr ,_{| |}b_rr ,_{| |r}cr . (ii) Incentre formula : The position vector of the incentre

of ∆ABC is aa bb cc

a b c

r r r

+ +

+ + .

(iii) Orthocentre formula : The position vector of the orthocentre of ∆ABC is

r r r

a A b B c C

A B C

tan tan tan

tan tan tan

+ +

+ +

(iv) Vector equation of a straight line passing through a fixed point with position vector _ar and parallel to a given vector br is r ar= + λr br.

(v) The vector equation of a line passing through two points with position vectors _ar and _br is

r r r r

r a= +λ

e j

b a− _.

(vi) Shortest distance between two parallel lines : Let l1 and l2 be two lines whose equations are l1 :

r r r

r a= ₁+λb₁ and l2 :

r r r

r a= ₂+µb₂ respectively. Then, shortest distance

PQ = b1 b_b2 _ba2 a1 b b a_{b b} a 1 2 1 2 2 1 1 2 × − × = − ×

c

h c

.

h

c

h

| | | |

shortest distance between two parallel lines : The shortest distance between the parallel lines r ar r= ₁ +λbr

and r ar r= ₂+µbr is given by d = | | | | r r r r a a b b 2 − 1 ×

c

h

_.

If the lines r ar r= ₁+λbr₁ and r ar = r₂+µbr₂ intersect, then the shortest distance between them is zero. Therefore, [b b a1 2

c

2−a1

h

] = 0

⇒ [

c

ar₂ −a b br r r₁

h

_{1 2] = 0} ⇒

c

ar₂ −ar₁

h e

. br₁ ×br₂

j

_{= 0.}

(vii) Vector equation of a plane normal to unit vector _nr and at a distance d from the origin is

r

r n. $ = d.

If _nr is not a unit vector, then to reduce the equation

r r

r n. = d to normal form we divide both sides by |_nr|

r nr d r d

(viii) The equation of the plane passing through a point having position vector _ar and parallel to _br and _cr is

r r r r

r a= +λb+µc or [r bcr r r] = [abcr r r], where λ and µ are scalars.

(ix) Vector equation of a plane passing through a point

r r r

abc is rr =

c

1− −s t a sbt c

h

r+ r +r

or r r r r r r rr b c c a a b.

e

× + × + ×

j

_{= [abc}r r r_].

(x) The equation of any plane through the intersection of planes r nr r. ₁ = d1 and r r r n. ₂ = d2 is r r r .

c

n₁ + λn₂

h

_{= d1 +} λd2, where λ is an arbitrary constant.

(xi) The perpendicular distance of a point having position vector ar from the plane r nr r. = d is given by p = | ._{| |} | r r r a n d n − .

(xii) An angle θ between the planes r nr r₁. ₁ =d₁ and r r r n2. 2 =d2 is given by cos θ = ± n n n11 n22 . | || |.

(xiii) The equation of the planes bisecting the angles between the planes r nr r1. 1 = d1

and r nr r2. 2 = d2 are | . | | | | . | | | r r r r r r r n d n r n d n 1 1 1 2 2 2 − ₌ −

(xiv) The plane r nr r. = d touches the sphere |r rr a− | = R,

if | ._{| |} | r r r a n d n − = R.

(xv) If the position vectors of the extremities of a diam- eter of a sphere are ar and br, then its equation is (_{r a}r r− ).(r br r− ) = 0 or |rr|2 – r a br r r.

e j

− +a br r. = 0.

In document Math Formula Sheet AIEEE (Page 83-90)